Case study (4 Marks)

Question 14 Marks

The management committee of a residential colony decided to award some of its members (say $x$ ) for honesty, some (say $y$ ) for helping others and some others (say $z$ ) for supervising the workers to kept the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. The sum of the number of awardees for honesty and supervision is twice the number of awardees for helping.

Q.1. $x+y+z= $ __________.
(A) 3 $\quad$ (B) 5
(C) 7 $\quad$ (D) 12
Q.2. $x-2 y=$ __________.
(A) $z$ $\quad$ (B) $-z$
(C) $2 z$$\quad$(D) $-2 z$
Q.3. The value of $z$ is __________.
(A) 3 $\quad$ (B) 4
(C) 5 $\quad$ (D) 6
Q.4. The value of $x+2 y=$ __________.
(A) 9 $\quad$ (B) 10
(C) 11$\quad$(D) 12

Answer

(1) (D) 12
Explanation:
$\begin{aligned}x+y+z & =12\quad \quad \ldots \ldots\text{(i)} \\2 x+3 y+3 z & =33\quad \quad \ldots \ldots\text{(ii)} \\x-2 y+z & =0 \quad \quad \ldots \ldots\text{(iii)}\end{aligned}$
$A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1\end{array}\right], B=\left[\begin{array}{c}12 \\ 33 \\ 0\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$
$\begin{aligned}|A| & =1(3+6)-1(2-3)+1(-4-3) \\ & =9+1-7 \\ & =3\end{aligned}$
$\begin{aligned} A^{-1} & =\frac{1}{|A|}(\operatorname{adj} A) \\ & =\frac{1}{3}\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right]\end{aligned}$
$\begin{aligned} X & =A^{-1} B \\ & =\frac{1}{3}\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right]\left[\begin{array}{c}12 \\ 33 \\ 0\end{array}\right]\end{aligned}$
$\begin{array}{l}=\frac{1}{3}\left[\begin{array}{c}9 \\ 12 \\ 15\end{array}\right] \\ =\left[\begin{array}{l}3 \\ 4 \\ 5\end{array}\right]\end{array}$
$\Rightarrow x=3,y=4,z=5$
$x+y+z=12\quad\quad\quad\quad\quad\text{[from (i)]}$
(2) (B) $-z$
Explanation: $x-2y=-z\quad\text{[from (iii)]}$
(3) (C) $5$
Explanation: $z=5$
(4) (C) $11$
Explanation: $x+2y=3+8=11$

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Question 24 Marks

A trust invested some money in two types of bonds. The first bond pays $10\%$ interest and second bond pays $12 \%$ interest the trust received ₹ $2800$ as interest. However if trust had interchanged money in bonds, they would have got ₹ $100$ less as interest.

Q.1. For the given information, the system of equations can be written as
(A)
$\begin{array}{l}5 x+6 y=135000 \\6 x+5 y=140000\end{array}$
(B)
$5 x+6 y=140000$
$6 x+5 y=135000$
(C)
$\begin{array}{l}5 x+5 y=140000 \\6 x+6 y=135000\end{array}$
(D)
$\begin{aligned}x+y & =140000 \\6 x+5 y & =135000\end{aligned}$
Q.2. If the given system of equations can be written as $A X=B$, then find $|A|$.
(A) -11
(B) 11
(C) -5
(D) -10
Q.3. The amount invested in first type of bond is
(A) ₹ 9000
(B) ₹ 12000
(C) ₹ 15000
(D) ₹ 10000
Q.4. The amount invested in first type of bond is
(A) ₹ 9000
(B) ₹ 12000
(C) ₹ 15000
(D) ₹ 10000

Answer

(1) (B)
$\begin{array}{l}5 x+6 y=140000 \\ 6 x+5 y=135000\end{array}$
Explanation: According to the questions:
$\left[\begin{array}{ll}x & y\end{array}\right]\left[\begin{array}{l}\frac{10}{100} \\ \frac{12}{100}\end{array}\right]=[2800]$
$\text{i.e.,}\quad \frac{10x}{100}+\frac{12y}{100}=2800$
$\Rightarrow 10x+12y=280000$
$\Rightarrow 5x+6y=140000\quad\ldots\text{(i)}$
$\text{Also,}\left[\begin{array}{ll}y & x\end{array}\right]\left[\begin{array}{l}\frac{10}{100} \\ \frac{12}{100}\end{array}\right]=[2700]$
$\text{i.e.,}\quad \frac{10y}{100}+\frac{12x}{100}=2700$
$\Rightarrow 10y+12x=270000$
$\Rightarrow 5y+6x=135000\quad\ldots\text{(ii)}$
(2) (A) $-11$
Explanation: By matrix method,
$\left[\begin{array}{ll}5 & 6 \\ 6 & 5\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}140000 \\ 135000\end{array}\right]$
$\text{where, }A=\left[\begin{array}{ll}5 & 6 \\ 6 & 5\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right], B=\left[\begin{array}{l}140000 \\ 135000\end{array}\right]$
It is of the form $AX=B$
$\Rightarrow X=A^{-1}B$
$|A|=\left|\begin{array}{ll}5 & 6 \\ 6 & 5\end{array}\right|$
$=25-36=-11\neq0$
(3) (D) ₹ $10000$
Explanation: Since $|A|\neq0,$ thus $A^{-1}$ exists, so the given system of equation has unique solution
$X=A^{-1}B$
Co-factors of $|A|$ are
$A_{11}=5,\quad A_{12}=-6,$
$A_{21}=-6, \quad A_{22}=5$
$\therefore \quad \operatorname{adj} A=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right]^T=\left|\begin{array}{cc}5 & -6 \\ -6 & 5\end{array}\right|$
$A^{-1}=\frac {1}{|A|}(\text{adj)}A$ and
$X=A^{-1}B$
$=\frac{1}{-11}\left[\begin{array}{cc}5 & -6 \\ -6 & 5\end{array}\right]\left[\begin{array}{l}140000 \\ 135000\end{array}\right]$
$=\frac{1}{-11}\left[\begin{array}{c}700000-810000 \\ -840000+675000\end{array}\right]$
$=\frac{1}{-11}\left[\begin{array}{l}-110000 \\ -165000\end{array}\right]$
$\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}10000 \\ 15000\end{array}\right]$
Thus, the amount invested in first type of bond is ₹ $x=$ ₹ $10000$
(4) (C) ₹ $15000$
Explanation: From the above, it clear that the amount invested in first type of bond is ₹ $y=$ ₹ $15000$

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Question 34 Marks

A shopkeeper has 3 varieties of pen 'A', 'B' and 'C'. Meenu purchase 1 pen of each variety for a total of ₹ 21 . Jeevan purchase 4 pens of 'A' variety, 3 pens say 'B' variety and 2 pens of 'C' variety for ₹ 60. While Shikha purchased 6 pens of 'A' variety, 2 pens of 'B' variety and 3 pens of 'C' variety for ₹ 70.

Q.1. For the given information, the system of equations can be written as:
(A)
$\begin{array}{r}x+y+z=21 \\4 x+3 y+2 z=60 \\6 x+2 y+3 z=70\end{array}$
(B)
$\begin{aligned}x+y+z & =60 \\4 x+3 y+2 z & =21 \\6 x+2 y+3 z & =70\end{aligned}$
(C)
$\begin{aligned} 4 x+3 y+2 z & =21 \\ x+y+z & =70 \\ 6 x+2 y+3 z & =60\end{aligned}$
(D)
$\begin{aligned} x+y+z & =70 \\ 4 x+3 y+2 z & =60 \\ 6 x+2 y+3 z & =21\end{aligned}$
Q.2. If the given system of equations can be written as $A X=B$, then find $|A|$.
(A) 10 $\quad$ (B) 5
(C) -5 $\quad$ (D) -10
Q.3. The cost of '$A$' variety pen is
(A) ₹ 4 $\quad$ (B) ₹ 5
(C) ₹ 8 $\quad$ (D) ₹ 2
Q.4. The cost of '$B$' variety pen is
(A) ₹ 4 $\quad$ (B) ₹ 5
(C) ₹ 8 $\quad$(D) ₹ 2

Answer

(1) (A)
$\begin{aligned} x+y+z & =21 \\ 4 x+3 y+2 z & =60 \\ 6 x+2 y+3 z & =70\end{aligned}$
Explanation: Let the cost of 'A' variety pen be $x$, the cost of 'B' variety pen be $y$, the cost of '$C$' variety pen be $z$.
According to the question,
$\begin{aligned}x+y+z & =21 \\4 x+3 y+2 z & =60 \\6 x+2 y+3 z & =70\end{aligned}$
(2) (C) -5
Explanation: The given system of the questions can be written as $A X=B$, i.e.
$\left[\begin{array}{lll}1 & 1 & 1 \\4 & 3 & 2 \\6 & 2 & 3\end{array}\right]\left[\begin{array}{l}x \\y \\z\end{array}\right]=\left[\begin{array}{l}21 \\60 \\70\end{array}\right]$
$\begin{aligned}\text{Now, }|A| & =\left|\begin{array}{lll}1 & 1 & 1 \\4 & 3 & 2 \\6 & 2 & 3\end{array}\right| \\& =1(9-4)-1(12-12) +1(8-18) \\& =5-10=-5 \neq 0\end{aligned}$
(3) (B) ₹ 5
Explanation: Since $|A| \neq 0$, thus $A^{-1}$ exists, so the given system of equation has unique solution $X=$ $A^{-1} B$
$\begin{array}{l}A_{11}=(9-4)=5 \\A_{12}=-(12-12)=0 \\A_{13}=(8-18)=-10 \\A_{21}=-(3-2)=-1 \\A_{22}=(3-6)=-3 \\A_{23}=-(2-6)=4 \\A_{31}=(2-3)=-1 \\A_{32}=-(2-4)=2 \\A_{33}=3-4=-1\end{array}$
$\operatorname{adj} A=\left[\begin{array}{ccc}5 & 0 & -10 \\ -1 & -3 & 4 \\ -1 & 2 & -1\end{array}\right]^T$
$\begin{aligned} & =\left[\begin{array}{ccc}5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1\end{array}\right] \\ A^{-1} & =\frac{1}{|A|}(\operatorname{adj} A)\end{aligned}$
$\text{Also,}\quad AX=B$
$\Rightarrow X=A^{-1}B$
$\Rightarrow A^{-1}=\frac{1}{|A|}(\text{adj } A)B$
$=\frac{1}{|A|}\left[\begin{array}{ccc}5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1\end{array}\right]\left[\begin{array}{l}21 \\ 60 \\ 70\end{array}\right]$
$=\frac{1}{-5}\left[\begin{array}{c}105-60-70 \\ 0-180+140 \\ -210+240-70\end{array}\right]$
$=\frac{1}{-5}\left[\begin{array}{l}-25 \\ -40 \\ -40\end{array}\right]$
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}5 \\ 8 \\ 8\end{array}\right]$
Hence, cost of '$A$' variety pen is ₹ $5.$
(4) (C) ₹ 8
Explanation: From the above, it clear that the cost of '$B$' variety pen is ₹ 8.

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Question 44 Marks

Two schools P and Q want to award their selected students on the values of the Discipline, Politeness and Punctuality. The school P wants to award ₹ $x$ each, ₹ $y$ each and ₹ $z$ each for the three respective values to its 3, 2 and 1 students with the total award money of ₹ $1,000.$ School Q wants spend ₹ $1,500$ to award its 4, 1 and 3 students on their respective values (by giving the same award money for the three values as before). The total amount of awards for one prize on each value is ₹ $600.$

Q.1. For the given information, the system of equations can be written as
(A)
$\begin{aligned}3 x+2 y+z & =1000 \\4 x+y+3 z & =1500 \\x+y+z & =600\end{aligned}$
(B)
$\begin{aligned}3 x+2 y+z & =1500 \\4 x+y+3 z & =1000 \\x+y+z & =600\end{aligned}$
(C)
$3 x+2 y+z=1000$
$\begin{aligned} x+y+z & =15000 \\ 4 x+y+3 z & =600\end{aligned}$
(D)
$\begin{aligned} x+2 y+z & =1000 \\ 4 x+y+3 z & =1500 \\ 3 x+2 y+z & =600\end{aligned}$
Q.2. If the given system of equations can be written as $A X=B$, then find $|A|$ :
(A) 10 $\quad$ (B) 5
(C) -5 $\quad$ (D) -10
Q.3. The award money for Discipline is:
(A) ₹ 100 $\quad$ (B) ₹ 200
(C) ₹ 300 $\quad$ (D) ₹ 400
Q.4. The award money for Politeness is:
(A) ₹ 100 $\quad$ (B) ₹ 200
(C) ₹ 300 $\quad$ (D) ₹ 400

Answer

(1) (A)
$\begin{aligned} 3 x+2 y+z & =1000 \\ 4 x+y+3 z & =1500 \\ x+y+z & =600\end{aligned}$
Explanation: According to the given conditions of the questions, we have
$\begin{aligned}3 x+2 y+z & =1000 \\4 x+y+3 z & =1500 \\x+y+z & =600\end{aligned}$
(2) (C) -5
Explanation: The given system of the questions can be written as $A X=B$, i.e.
$\left[\begin{array}{lll}3 & 2 & 1 \\4 & 1 & 3 \\1 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\y \\z\end{array}\right]=\left[\begin{array}{c}1000 \\1500 \\600\end{array}\right]$
$\begin{aligned}\text{Now, }|A| & =3(1-3)-2(4-3)+1(4-1) \\& =-6-2+3=-5\end{aligned}$
(3) (A) ₹ 100
Explanation: Since $|A| \neq 0$, thus $A^{-1}$ exists, so the given system of equation has unique solution $X=$ $A^{-1} B$
$\begin{array}{llll}\text { Hence, } & A_{11}=-2, & A_{12}=-1, & A_{13}=3 \\ & A_{21}=-1, & A_{22}=2, & A_{23}=-1 \\ \text { And } & A_{31}=5, & A_{32}=-5, & A_{33}=-5\end{array}$
$\begin{array}{ll}\therefore & \quad \operatorname{adj} A=\left[\begin{array}{ccc}-2 & -1 & 3 \\ -1 & 2 & -1 \\ 5 & -5 & -5\end{array}\right]=\left[\begin{array}{ccc}-2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5\end{array}\right] \\ \therefore & A^{-1}=\frac{1}{|A|} \text { adj } A=\frac{1}{-5}\left[\begin{array}{ccc}-2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5\end{array}\right]\end{array}$
$\begin{aligned} \therefore \quad X & =A^{-1} B \\ & =\frac{1}{-5}\left[\begin{array}{ccc}-2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5\end{array}\right]\left[\begin{array}{c}1000 \\ 1500 \\ 600\end{array}\right]\end{aligned}$
$\therefore \quad\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-5}\left[\begin{array}{c}-500 \\ -1000 \\ -1500\end{array}\right]=\left[\begin{array}{l}100 \\ 200 \\ 300\end{array}\right]$
(4) (B) ₹ 200
Explanation: From the above, it clear that the award money for Politeness is ₹ 200.

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Applied Maths Case study (4 Marks)

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